Find the equation of the circle which is tangential to the axis (2,0) and touches the line y=6
Unsure of workings?
Answers are; A, (x+2)^2 + (y7)^2 = 26 B, 2x y = 5 C, k = 0
Cheers Neils=)
Find the equation of the circle which is tangential to the axis (2,0) and touches the line y=6
Unsure of workings?
Answers are; A, (x+2)^2 + (y7)^2 = 26 B, 2x y = 5 C, k = 0
Cheers Neils=)
Thanks Grandad.(Happy) That was form my text book, looks like a missprint maybe... I will as my teacher.
Do you mean "tangent to the xaxis at (2,0)"? (2,0) is a point not a line.
If so, then both given tangent lines are horizontal and the center of the circle is at (2, 3).
There are no "A" or "B" in your question!Quote:
Unsure of workings?
Answers are; A, (x+2)^2 + (y7)^2 = 26 B, 2x y = 5 C, k = 0
Cheers Neils=)
Hello, Neils!
Did you make a sketch?Quote:
Find the equation of the circle tangent to the xaxis at (2,0) and tangent to the line y=6
Answers are:
$\displaystyle (A);\;(x+2)^2 + (y7)^2 \:=\:26\qquad (B)\;\;2x y \:=\:5 \qquad (C)\;\;k \,=\, 0 $
These are great answers . . . What were the questions?
Code:
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 (2,0)

Any idea where the center is and what the radius is?